a) \(\sqrt{24+8\sqrt{5}}+\sqrt{9-4\sqrt{5}}\)

\(=2\sqrt{5}+2+\sqrt{5}-2\)

\(=3\sqrt{5}\)

b) \(\sqrt{17-12\sqrt{2}}+\sqrt{9+4\sqrt{2}}\)

\(=3-2\sqrt{2}+2\sqrt{2}-1\)

=2

c) \(\sqrt{6-4\sqrt{2}}+\sqrt{22-12\sqrt{2}}\)

\(=2-\sqrt{2}+3\sqrt{2}-2\)

\(=2\sqrt{2}\)

cho cách làm dạng bài này luôn. Chỗ nào chưa hiểu thì nói tớ sẽ giải thích thêm (cần góp ý để hoàn thiện thêm phần hướng dẫn đó mà. Cảm ơn cậu).

Phương Nam Phim (à quên, Từ Hạ) hân hạnh giới thiệu bộ phim...

a,\(\sqrt{24+8\sqrt{5}}+\sqrt{9-4\sqrt{5}}=\sqrt{2^2+2\cdot2\cdot\left(2\sqrt{5}\right)+\left(2\sqrt{5}\right)^2}\) \(+\sqrt{\left(\sqrt{5}\right)^2-2\cdot2\sqrt{5}+2^2}=\sqrt{\left(2+2\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}-2\right)^2}\)=\(2+2\sqrt{5}+\sqrt{5}-2=3\sqrt{5}\) 

b,\(\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(2\sqrt{2}+1\right)^2}=3-2\sqrt{2}+2\sqrt{2}+1=4\)

c,\(\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(3\sqrt{2}-2\right)^2}=2-\sqrt{2}+3\sqrt{2}-2=2\sqrt{2}\)

câu b với câu c giải thích ra dùm e đc kh ạ?

Lời giải:
a.

\(=\sqrt{5+2.2\sqrt{5}+2^2}-\sqrt{5-2.2\sqrt{5}+2^2}\)

$=\sqrt{(\sqrt{5}+2)^2}-\sqrt{(\sqrt{5}-2)^2}$

$=|\sqrt{5}+2|-|\sqrt{5}-2|=(\sqrt{5}+2)-(\sqrt{5}-2)=4$

b.

$=\sqrt{3-2.3\sqrt{3}+3^2}+\sqrt{3+2.3.\sqrt{3}+3^2}$

$=\sqrt{(\sqrt{3}-3)^2}+\sqrt{(\sqrt{3}+3)^2}$

$=|\sqrt{3}-3|+|\sqrt{3}+3|$

$=(3-\sqrt{3})+(\sqrt{3}+3)=6$

c.

$=\sqrt{2+2.3\sqrt{2}+3^2}-\sqrt{2-2.3\sqrt{2}+3^2}$

$=\sqrt{(\sqrt{2}+3)^2}-\sqrt{(\sqrt{2}-3)^2}$
$=|\sqrt{2}+3|-|\sqrt{2}-3|$

$=(\sqrt{2}+3)-(3-\sqrt{2})=2\sqrt{2}$

a,\(\sqrt{\left(\sqrt{3}-1\right)^2}\) \(+\sqrt{\left(\sqrt{3}+1\right)^2}=2\sqrt{3}\)

b. \(\sqrt{\left(2\sqrt{5}+2\right)^2}+\sqrt{\left(\sqrt{5}-2\right)^2}=3\sqrt{5}\)

c,\(\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(2\sqrt{2}+1\right)^2}=4\)

d.\(\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(3\sqrt{2}-2\right)^2}=2\sqrt{2}\)

a: \(=\dfrac{\sqrt{3}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}-3\sqrt{3}+\dfrac{\sqrt{3}\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{3}+\sqrt{2}}\)

\(=\sqrt{3}-3\sqrt{3}+\sqrt{3}=-\sqrt{3}\)

b: \(=\left(\left(2-2\sqrt{5}\right)\left(\sqrt{5}+2\right)+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\)

\(=\left(2\sqrt{5}+4-10-4\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\)

\(=\left(-2\sqrt{5}+\sqrt{3}-6\right)\left(\sqrt{5}-\sqrt{3}\right)\)

\(=-20+2\sqrt{15}+\sqrt{15}-3-6\sqrt{5}+6\sqrt{3}\)

\(=-23+3\sqrt{15}-6\sqrt{5}+6\sqrt{3}\)

a,

\(\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}\\ =\sqrt{3-2\sqrt{3}+1}+\sqrt{3+2\sqrt{3}+1}\\ =\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\\ =\left|\sqrt{3}-1\right|+\left|\sqrt{3}+1\right|\\ =\sqrt{3}-1+\sqrt{3}+1\\ =2\sqrt{3}\)

b,

\(\sqrt{24+8\sqrt{5}}+\sqrt{9-4\sqrt{5}}\\ =\sqrt{24+4\cdot\sqrt{4}\cdot\sqrt{5}}+\sqrt{9-4\sqrt{5}}\\ =\sqrt{24+4\sqrt{20}}+\sqrt{9-4\sqrt{5}}\\ =\sqrt{20+4\sqrt{20}+4}+\sqrt{5-4\sqrt{5}+4}\\ =\sqrt{\left(\sqrt{20}+4\right)^2}+\sqrt{\left(\sqrt{5}-2\right)^2}\\ =\left|\sqrt{20}+4\right|+\left|\sqrt{5}-2\right|\\ =\sqrt{20}+4+\sqrt{5}-2\\ =2+2\sqrt{5}+\sqrt{5}\\ =2+3\sqrt{5}\)

c,

\(\sqrt{17-12\sqrt{2}}+\sqrt{9+4\sqrt{2}}\\ =\sqrt{17-6\cdot\sqrt{4}\cdot\sqrt{2}}+\sqrt{9+2\cdot\sqrt{4}\cdot\sqrt{2}}\\ =\sqrt{17-6\sqrt{8}}+\sqrt{9+2\sqrt{8}}\\ =\sqrt{9-6\sqrt{8}+8}+\sqrt{8+2\sqrt{8}+1}\\ =\sqrt{\left(3-\sqrt{8}\right)^2}+\sqrt{\left(\sqrt{8}+1\right)^2}\\ =\left|3-\sqrt{8}\right|+\left|\sqrt{8}+1\right|\\ =3-\sqrt{8}+\sqrt{8}+1\\ =4\)

d,

\(\sqrt{6-4\sqrt{2}}+\sqrt{22-12\sqrt{2}}\\ =\sqrt{6-4\sqrt{2}}+\sqrt{22-4\cdot\sqrt{9}\cdot\sqrt{2}}\\ =\sqrt{6-4\sqrt{2}}+\sqrt{22-4\sqrt{18}}\\ =\sqrt{4-4\sqrt{2}+2}+\sqrt{18-4\sqrt{18}+4}\\ =\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{18}-2\right)^2}\\ =\left|2-\sqrt{2}\right|+\left|\sqrt{18}-2\right|\\ =2-\sqrt{2}+\sqrt{18}-2\\ =-\sqrt{2}+\sqrt{18}\\ =-\sqrt{2}+3\sqrt{2}\\ =2\sqrt{2}\)

b,

\(\sqrt{24+8\sqrt{5}}+\sqrt{9-4\sqrt{5}}\\ =\sqrt{24+4\cdot\sqrt{4}\cdot\sqrt{5}}+\sqrt{9-4\sqrt{5}}\\ =\sqrt{24+4\sqrt{20}}+\sqrt{9-4\sqrt{5}}\\ =\sqrt{20+4\sqrt{20}+4}+\sqrt{5-4\sqrt{5}+4}\\ =\sqrt{\left(\sqrt{20}+2\right)^2}+\sqrt{\left(\sqrt{5}-2\right)^2}\\ =\left|\sqrt{20}+2\right|+\left|\sqrt{5}-2\right|\\ =\sqrt{20}+2+\sqrt{5}-2\\ =\sqrt{20}+\sqrt{5}\\ =\sqrt{4}\cdot\sqrt{5}+\sqrt{5}\\ =2\sqrt{5}+\sqrt{5}\\ =3\sqrt{5} \)

a: Ta có: \(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{5}{\sqrt{5}}\right):\dfrac{1}{\sqrt{5}-\sqrt{2}}\)

\(=-\left(\sqrt{5}+\sqrt{2}\right)\left(\sqrt{5}-\sqrt{2}\right)\)

=-5+2

=-3

b) \(\sqrt{7-2\sqrt{10}}-\sqrt{7+2\sqrt{10}}\)

\(=\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{2}+2}-\sqrt{5+2\cdot\sqrt{5}\cdot\sqrt{2}+2}\)

\(=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}\)

\(=\left|\sqrt{5}-\sqrt{2}\right|-\left|\sqrt{5}+\sqrt{2}\right|\)

\(=\sqrt{5}-\sqrt{2}-\sqrt{5}-\sqrt{2}\) (vì \(\sqrt{5}\ge\sqrt{2}\)

=0

c) \(\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}\)

\(=\sqrt{3-2\sqrt{3}+1}+\sqrt{3+2\sqrt{3}+1}\)

\(=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\)

\(=\left|\sqrt{3}-1\right|+\left|\sqrt{3}+1\right|\)

\(=\sqrt{3}-1+\sqrt{3+1}\) (vì \(\sqrt{3}\ge1\))

\(=2\sqrt{3}\)

a)\(\sqrt{5+2\sqrt{6}}-\sqrt{5+2\sqrt{6}}\)

\(=\sqrt{3+2\cdot\sqrt{3}\cdot\sqrt{2}+2}-\sqrt{3-2\cdot\sqrt{3}\cdot\sqrt{2}+2}\)

\(=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)

\(=\left|\sqrt{3}+\sqrt{2}\right|-\left|\sqrt{3}-\sqrt{2}\right|\)

\(=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}\) (vì \(\sqrt{3}\ge\sqrt{2}\))

=0